Rational Root Theorem – Does It Guarantee Polynomial Irreducibility?

abstract-algebrairreducible-polynomialspolynomialsroots

I know that the Eisenstein's criterion can guarantee it (when applicable), but what about this?

Also, is there another test that you can use to test irreducibility besides these two? For example, $x^4+2x^2+49$, over the rationals?

Best Answer

With respect to your specific question (and quartics in general), the RRT doesn't solve the problem but it can help. Once you verify that your quartic has no roots, the only way it can factor is as a product of quadratic polynomials. Additionally, you can guarantee that the quadratic equations are monic and have integer coefficients. Observing that the constant terms multiply to $49$, we have four options:

$$x^4 + 2x^2 + 49 = (x^2 + ax + 7)(x^2 + bx + 7)$$

$$x^4 + 2x^2 + 49 = (x^2 + ax - 7)(x^2 + bx - 7)$$

$$x^4 + 2x^2 + 49 = (x^2 + ax + 1)(x^2 + bx + 49)$$

$$x^4 + 2x^2 + 49 = (x^2 + ax -1)(x^2 + bx - 49)$$

The vanishing of the cubic term gives $a = -b$ in each scenario, and the vanishing of the $x$ term eliminates scenarios 3 and 4. Comparing the $x^2$ terms in the first two scenarios gives:

$$14 - a^2 = 2$$

$$-14 - a^2 = 2$$

And it is easily seen that neither of these have solutions in the integers.

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